(Introduction to Probability, Blitzstein and Nwang, p.128)
People are arriving at a party one at a time. While waiting for more people to arrive they entertain themselves by comparing their birthdays. Let X be the number of people needed to obtain a birthday match, i.e., before person X arrives there are no two people with the same birthday, but when person X arrives there is a match. Find the PMF of X.
The CDF is $P(X\leq k) = 1 – \frac{\binom{365}{k}k!}{365^k}$, so the can obtain the PMF by
\begin{align}
P(X=k) &= P(X\leq k) – P(X\leq k-1)\\
&= \left( 1 – \frac{\binom{365}{k}k!}{365^k} \right) – \left( 1 – \frac{\binom{365}{k-1}(k-1)!}{365^{k-1}} \right) \\
&= \frac{(k-1)}{365^k} \binom{365}{k-1} (k-1)! \\
&=(k-1)*\left(1-P(X \leq k-1 \right)) \\
&= (k-1) * P(X > k-1)
\end{align}
Is this correct? How do you interpret the results? How to arrive at the PMF without using the CDF?
Best Answer
There is an easier way, which lead to the correct answer.
You want $P(k)$, the probability that the first match occurs at person $k$. For that to happen, the previous $k-1$ people must have no matches, and $X$ must match one of them.
The probability of $k-1$ people to have no matches is $$ \frac{N}{N} \frac{N-1}{N} \frac{N-2}{N} \cdots \frac{N-k+2}{N} = \frac{N!}{(N-k+1)! N^{k-1}} $$ with $N = 365$. Then the probability of a match on the k-th person is $$ P(k) = \frac{N!}{(N-k+1)! N^{k-1}} \frac{k-1}{N} = (k-1) \frac{N!}{(N-k+1)! N^{k}} $$
The expression you give is equivalent to $(k-1) \frac{N!}{(N-k+1)! N^{k}}$. But this derivation did not use the CDF.